标签：java org following term tree eve unit OWIN tco
110. Balanced Binary Tree
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ 9 20
/ 15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ 2 2
/ 3 3
/ 4 4
Return false.
package leetcode.easy;
public class BalancedBinaryTree {
@org.junit.Test
public void test1() {
TreeNode tn11 = new TreeNode(3);
TreeNode tn21 = new TreeNode(9);
TreeNode tn22 = new TreeNode(20);
TreeNode tn33 = new TreeNode(15);
TreeNode tn34 = new TreeNode(7);
tn11.left = tn21;
tn11.right = tn22;
tn21.left = null;
tn21.right = null;
tn22.left = tn33;
tn22.right = tn34;
tn33.left = null;
tn33.right = null;
tn34.left = null;
tn34.right = null;
System.out.print(isBalanced(tn11));
}
@org.junit.Test
public void test2() {
TreeNode tn11 = new TreeNode(1);
TreeNode tn21 = new TreeNode(2);
TreeNode tn22 = new TreeNode(2);
TreeNode tn31 = new TreeNode(3);
TreeNode tn32 = new TreeNode(3);
TreeNode tn41 = new TreeNode(4);
TreeNode tn42 = new TreeNode(4);
tn11.left = tn21;
tn11.right = tn22;
tn21.left = tn31;
tn21.right = tn32;
tn22.left = null;
tn22.right = null;
tn31.left = tn41;
tn31.right = tn42;
tn32.left = null;
tn32.right = null;
tn41.left = null;
tn41.right = null;
tn42.left = null;
tn42.right = null;
System.out.print(isBalanced(tn11));
}
public boolean isBalanced(TreeNode root) {
if (null == root) {
return true;
} else {
return (Math.abs(maxDepth(root.left) - maxDepth(root.right)) <= 1) && isBalanced(root.left)
&& isBalanced(root.right);
}
}
private static int maxDepth(TreeNode root) {
if (null == root) {
return 0;
} else {
return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
}
}
}
LeetCode_110. Balanced Binary Tree
标签：java org following term tree eve unit OWIN tco
原文地址：https://www.cnblogs.com/denggelin/p/11616872.html
